Divide using synthetic division $$ ( 3x^{4}+6x^{3}-2x+4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&6&0&-2&4\\& & -6& 0& 0& \color{black}{4} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+6x^{3}-2x+4 }{ x+2 } = \color{blue}{3x^{3}-2} ~+~ \frac{ \color{red}{ 8 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&6&0&-2&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&6&0&-2&4\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&6&0&-2&4\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 6 }&0&-2&4\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&6&0&-2&4\\& & -6& \color{blue}{0} & & \\ \hline &3&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&6&\color{orangered}{ 0 }&-2&4\\& & -6& \color{orangered}{0} & & \\ \hline &3&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&6&0&-2&4\\& & -6& 0& \color{blue}{0} & \\ \hline &3&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-2&3&6&0&\color{orangered}{ -2 }&4\\& & -6& 0& \color{orangered}{0} & \\ \hline &3&0&0&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&6&0&-2&4\\& & -6& 0& 0& \color{blue}{4} \\ \hline &3&0&0&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 4 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&3&6&0&-2&\color{orangered}{ 4 }\\& & -6& 0& 0& \color{orangered}{4} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-2 } $ with a remainder of $ \color{red}{ 8 } $.